# How to show $\frac{\sinθ +\tanθ}{2}>\theta$ using geometry? [duplicate]

I could show in calculus where $0<θ<\frac{\pi}{2}$ like this

$$\frac{d(θ)}{dθ}=1<\frac{1}{2}(\cosθ+\sec^2θ)$$

But I curioused about would it could prove by geometric way likewise trigonometric identities. I try to show with the area of circular sector but I failed...

So this is question:

How to show $$\frac{\sinθ +\tanθ}{2}>\theta$$ where $0<\theta<\frac{\pi}{2}$ in geometric way?

## marked as duplicate by Namaste, Blue geometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 23 '18 at 16:39

• Recall that $\theta$ is the length of the arc of radius $\theta$. This may be of help to prove the inequality geometrically on the unit circle – b00n heT Aug 23 '18 at 15:19
• Please fix the geomatic/geomatric typos. – Yves Daoust Aug 23 '18 at 15:23
• @YvesDaoust I fixed it – user366725 Aug 23 '18 at 15:25
• @user366725: I am afraid not. – Yves Daoust Aug 23 '18 at 15:26
• If you are interested in more calculus-based proofs, you can check Jack's and Martin's answer here: math.stackexchange.com/questions/2196624/… – Botond Aug 23 '18 at 15:33

Here are some thoughts:

Approach $1a$

The blue region is a unit arc with an angle $\theta$. Let the length of $\overline {IH}$ be $\displaystyle L=\frac{\sin\theta+\tan\theta}2$.

The area of $\triangle AIH$ is given by $$S=\frac12 L\left(\frac L{\tan\theta}\right)=\frac{L^2}{2\tan\theta}\tag{1}$$

Notice the area of $\triangle ADG> \text{arc } ACG\implies\tan\theta>\theta\tag 2$

By $(1), (2)$

$$S<\frac{L^2}{2\theta}$$

Approach 1b

Notice arc $CJG$ has the length $\theta$. Then to prove $\theta<L$ is equivalent to prove the length of arc $CJG$ is shorter than the length of line $\overline {IH}$.

• To be honest, I don't think my attempt deserves to be an accepted answer. It is incomplete, plus it has a better and complete solution in the previous question. – Mythomorphic Aug 26 '18 at 16:07